Find the solutions to $z^4 = -4.$  Enter the solutions, separated by commas.
Let $z = x + yi,$ where $x$ and $y$ are real.  Then
\[(x + yi)^4 = x^4 + 4ix^3 y - 6x^2 y^2 - 4ixy^3 + y^4 = -4.\]Equating real and imaginary parts, we get
\begin{align*}
x^4 - 6x^2 y^2 + y^4 &= -4, \\
4x^3 y - 4xy^3 &= 0.
\end{align*}From the equation $4x^3 y - 4xy^3 = 0,$ $4xy(x^2 - y^2) = 0.$  If $x = 0,$ then $y^4 = -4,$ which has no solutions.  If $y = 0,$ then $x^4 = -4,$ which has no solutions.  Otherwise, $x^2 = y^2.$

Then the first equation becomes $-4x^4 = -4,$ so $x^4 = 1.$  Hence, $x = 1$ or $x = -1.$  In either case, $x^2 = 1,$ so $y^2 = 1,$ and $y = \pm 1.$  Therefore, the solutions are $\boxed{1 + i, 1 - i, -1 + i, -1 - i}.$